Answer
$$\int\left[e^{t} \mathbf{i}+\sin t \mathbf{j}+\cos t \mathbf{k}\right] d t=e^{t} \mathbf{i}-\cos t \mathbf{j}+\sin t \mathbf{k}+\mathbf{C}$$
Work Step by Step
Given $$\int\left[e^{t} \mathbf{i}+\sin t \mathbf{j}+\cos t \mathbf{k}\right] d t $$
Integrating on a component-by-component basis produces
\begin{align}
I&=\int\left[e^{t} \mathbf{i}+\sin t \mathbf{j}+\cos t \mathbf{k}\right] d t=e^{t} \mathbf{i}-\cos t \mathbf{j}+\sin t \mathbf{k}+\mathbf{C}
\end{align}
